MCQ
If $\text{x}-\frac{1}{\text{x}}=\frac{15}{4},$ than $\Big(\text{x}+\frac{1}{\text{x}}\Big)=$
- A$\frac{1}{4}$
- B$\frac{13}{4}$
- C$4$
- ✓$\frac{17}{4}$
✓
Answer
Correct option: D.
$\frac{17}{4}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\frac{15}{4}$
Now, $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\Big(\frac{15}{4}\Big)^2$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)-2\times\text{x}\times\frac{1}{\text{x}}=\frac{225}{16}$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)=\frac{225}{16}+2$
$\Rightarrow(\text{x})^2+\Big(\frac{1}{\text{x}^2}\Big)=\frac{257}{16}$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{257}{16}+2\times\text{x}\times\frac{1}{\text{x}}$
$=\frac{257}{16}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\frac{257+32}{16}=\frac{289}{16}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)=\sqrt{\frac{289}{16}}=\frac{17}{4}$
Now, $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\Big(\frac{15}{4}\Big)^2$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)-2\times\text{x}\times\frac{1}{\text{x}}=\frac{225}{16}$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)=\frac{225}{16}+2$
$\Rightarrow(\text{x})^2+\Big(\frac{1}{\text{x}^2}\Big)=\frac{257}{16}$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{257}{16}+2\times\text{x}\times\frac{1}{\text{x}}$
$=\frac{257}{16}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\frac{257+32}{16}=\frac{289}{16}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)=\sqrt{\frac{289}{16}}=\frac{17}{4}$
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